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Let $(X,d)$ be a metric space such that $x_{1},x_{2}\in X$ and $r_{1},r_{2}>0$ and let assume that $p\in B(x_{1},r_{1})\cap B(x_{2},r_{2})$. let $r=min\{r_{1}-d(p,x_{1}),r_{2}-d(p,x_{2})\}$ prove:

$$B(p,r)\subseteq B(x_{1},r_{1})\cap B(x_{2},r_{2})$$

Let $y\in B(p,r)$

$d(p,y)\leq d(p,x_{1})+d(x_{1},y)<r_{1}+d(x_{1},y)$ and $d(p,y)\leq d(p,x_{2})+d(x_{2},y)<r_{2}+d(x_{2},y)$

$d(p,y)<r_{1}+d(x_{1},y)$ and $d(p,y)<r_{2}+d(x_{2},y)$

$d(p,y)-d(x_{1},y)<r_{1}$ and $d(p,y)-d(x_{2},y)<r_{2}$

How should I continue?

Community
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gbox
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3 Answers3

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Hint: for every ball (skipping indexing): $$ d(x,y)\le d(x,p)+d(p,y)\le d(p,x)+r-d(p,x)=r. $$

A.Γ.
  • 29,518
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If $y\in B(p, r)$, then $$d(p, y)<r\leq r_1-d(p, x_1)\Rightarrow d(p, y)+d(p, x_1)< r_1\Rightarrow d(y, x_1)< r_1\Rightarrow y\in B(x_1, r_1)$$ because $d(y, x_1)\leq d(p, y)+d(p, x_1)$ by triangle inequality. Similarly $y\in B(x_2, r_2)$.

Prajwal Kansakar
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Let $y\in B(p,r)$. So $d(p,y)<r$. Then $d(p,y)<r\leq r_1-d(p,x_1)$ and $d(p,y)<r\leq r_2-d(p,x_2)$. So $d(p,y)+d(p,x_1)< r_1$ and $d(p,y)+d(p,x_2)< r_2$. Therefore $d(y,x_1)< r_1$ and $d(y,x_2)< r_2$. Hence $y\in B(x_{1},r_{1})\cap B(x_{2},r_{2})$

S.S.Danyal
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